$f(t) = -6t^{2}-6t-5-3(g(t))$ $g(t) = t$ $h(t) = -4t^{2}+4t+g(t)$ $ f(h(-1)) = {?} $
Explanation: First, let's solve for the value of the inner function, $h(-1)$ . Then we'll know what to plug into the outer function. $h(-1) = -4(-1)^{2}+(4)(-1)+g(-1)$ To solve for the value of $h$ , we need to solve for the value of $g(-1)$ $g(-1) = -1$ $g(-1) = -1$ That means $h(-1) = -4(-1)^{2}+(4)(-1)-1$ $h(-1) = -9$ Now we know that $h(-1) = -9$ . Let's solve for $f(h(-1))$ , which is $f(-9)$ $f(-9) = -6(-9)^{2}+(-6)(-9)-5-3(g(-9))$ To solve for the value of $f$ , we need to solve for the value of $g(-9)$ $g(-9) = -9$ $g(-9) = -9$ That means $f(-9) = -6(-9)^{2}+(-6)(-9)-5+(-3)(-9)$ $f(-9) = -410$